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EE 134 Homework 2 Spring 2023 At V oc , net current I net = 0 I = I L − I 0 [ exp ( qv 2 kT ) − 1 ] = 0 I 0 = 400 x 10 -12 A Since the diode is operating at 100% efficiency, each photon excites each electron and thus producing the equivalent current. I L = ϕ × A×q I L = 2.25 × 10 21 × 1 × 10 − 4 × 1.6 × 10 − 19 I L = 0.036 A I L − I 0 [ exp ( qv 2 kT ) − 1 ] = 0 At 300K, we get value of V = 0.95V Power = V x I P = V × ( I L − I 0 [ exp ( qV 2 kT ) − 1 ] ) P = V × ( 0.036 − 400 × 10 − 12 × [ exp ( qV 2 kT ) − 1 ] ) We plot a graph of P vs V to get the maximum power from the graph. We get maximum power at V ~ 0.8 V This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

For part b, Maximum Power = 0.027 W P = V 2 R 0.027 = 0.8 2 R Resistance = 23.7 ohm Incoming power = 1000 W m 2 × 1 × 10 − 4 m 2 = 0.1 W Power Efficiency = 0.027 0.1 = 27% For 1000 suns, I L = 0.036 A x 1000 = 36A V oc = 1.3 V We get maximum power as 39.36 W at 1.14 V Incoming power = 0.1 W x 1000 = 100W Power Efficiency = 39.36 100 = 39.36% This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

V d = V p = I p R p = V + I R s I p = V + I R s R p I = I L − I d − I p I d = I 0 [ exp ( q ( V + I R s ) mkT ) − 1 ] I = I L − I 0 [ exp ( q ( V + I R s ) mkT ) − 1 ] − V + I R s R p For V oc ; I = 0 0 = I L − I 0 [ exp ( qV oc mkT ) − 1 ] − V oc R p For I sc ; V = 0 I sc = I sc − I 0 [ exp ( q I sc R S mkT ) − 1 ] − I sc R s R p This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

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Condition I sc V oc Rs→0 and Rp→∞ I = I L − I 0 [ exp ( qV mkT ) − 1 ] I sc = I L I = I L − I 0 [ exp ( qV mkT ) − 1 ] I L = I 0 [ exp ( qV oc mkT ) − 1 ] Rs→0, ±nite Rp I = I L − I 0 [ exp ( qV mkT ) − 1 ] − V R I sc = I L I = I L − I 0 [ exp ( qV mkT ) − 1 ] − V R I L = I 0 [ exp ( qV oc mkT ) − 1 ] + V oc R p Rs nonzero, Rp→∞ I = I L − I 0 [ exp ( q ( V + I R s ) mkT ) − I = I sc − I 0 [ exp ( qI R S mkT ) − 1 ] I = I L − I 0 [ exp ( q ( V + I R s ) mkT ) − I L = I 0 [ exp ( qV oc mkT ) − 1 ] Parasitic shunt and series resistances are caused by the defects and resistance present in the components and materials used in the circuit. These resistances lead to decreased efficiency and performance of the circuit. The series resistance is in series with the circuit and causes a drop in voltage thus leading to power dissipation. The shunt resistance is in parallel with the diode and leads to a reduction in the overall circuit current. This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

I = I L − I 0 [ exp ( q ( V + I R s ) mkT ) − 1 ] − V + I R s R p I = 1 − 10 × 10 − 6 [ exp ( 1.6 × 10 − 19 × ( V + 0.1 ×I ) 8.284 × 10 − 21 ) − 1 ] − V + 0.1 × I 100 Condition Equation Graph Isc Voc This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

Rs→0 and Rp→∞ I = 1 − 10 × 10 − 6 [ exp ( 1.6 × 10 8.284 × Rs→0, finite Rp I = 1 − 10 × 10 − 6 [ exp ( 1.6 × 10 8.284 × Rs nonzero, Rp→∞ I = 1 − 10 × 10 − 6 [ exp ( 1.6 × 10 8.  The parasitic resistance has no effect on the short circuit current while it reduces the open circuit voltage because it is in parallel to the diode.  The series resistance, however, does not change the open circuit voltage but reduces the short circuit current because it is in series with the diode circuit. I = 1 − 50 × 10 − 6 [ exp ( 1.6 × 10 − 19 × ( V + 0.1 ×I ) 8.284 × 10 − 21 ) − 1 ] − V + 0.1 × I 100 As I o increases to 50 uA, there is no change in I sc but V oc changes to 0.512 V. Also, the curve moved closer to the y-axis meaning that the fill factor has reduced. Voc Isc Isc Voc Isc Voc This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

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3. (10 pts) Solar radiation, bandgap, and absorption a. (1 pts) What are the differences between the blackbody spectrum at 5900K and AM0 spectrum? What are the causes of the differences? Please cite your sources.  A blackbody is an object that absorbs all electromagnetic radiation incident on it. Radiation emitted by a blackbody at 5900K is its spectrum.  The AM0 spectrum is the spectrum of solar radiation at the top of the Earth's atmosphere, when the sun is at its highest point. It was measured with no air between the sun and the receiver.  The sun is not an ideal blackbody. So, the difference arises due to absorption of the solar radiations in the atmosphere. Due to this, AM0 spectrum has a reduced intensity in the UV region. Sources: Massiot, Inès. (2013). Design and fabrication of nanostructures for light-trapping in ultra-thin solar cells. 10.13140/RG.2.1.2731.0167. https://www2.pvlighthouse.com.au/resources/courses/altermatt/The%20Solar %20Spectrum/The%20extraterrestrial%20(AM0)%20solar%20spectrum.aspx b. (1 pts) What are the differences between the blackbody spectrum at AM0 and AM1.5 spectrum? What are the causes of the differences? Please cite your sources.  The AM0 spectrum is the spectrum of solar radiation at the top of the Earth's atmosphere, when the sun is at its highest point. It was measured with no air between the sun and the receiver. The standard spectrum of AM1.5 corresponds to the sun being at an angle of elevation of 42°.  The differences are due to absorption or scattering of the incoming radiation in the Earth's atmosphere. The AM1.5 spectrum has a lower intensity in the UV and visible region as compared to AM0 spectrum. This is because the Earth's atmosphere absorbs and scatters more UV and visible light and less near-infrared light. Sources: Kruse, Olaf & Rupprecht, Jens & Mussgnug, Jan & Dismukes, Gerard & Hankamer, Ben. (2006). Photosynthesis: A blueprint for solar energy capture and biohydrogen production Isc Voc This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

technologies. Photochemical & photobiological sciences: Official journal of the European Photochemistry Association and the European Society for Photobiology. 4. 957-70. 10.1039/b506923h. c. (1 pts) Why do we use the AM1.5 spectrum to characterize cells? What is a typical approximation for the integrated irradiance of the AM1.5 spectrum (in W/m2)? Please cite your sources. The AM1.5 spectrum is used to characterize cells because it is a good representation yearly average irradiance in the temperate latitudes where there are many large population centres. The integrated irradiance of the AM1.5 spectrum is approximated as 1KW/m 2 . Source: https://www.ossila.com/en-us/pages/standard-solar-spectrum d. (1 pts) What is the band-gap energy of crystalline Silicon and GaAs?  Eg of crystalline silicon is ~ 1.1 eV.  Eg of gallium arsenide is ~ 1.4 eV. e. (1 pts) What are the corresponding wavelengths at these energies?  Wavelength of crystalline silicon is ~ 1130 nm.  Wavelength of gallium arsenide is ~ 887 nm. f. (1 pts) What is the optical absorption length at band-edge wavelength for crystalline Silicon and GaAs?  Optical Absorption Length of crystalline silicon is ~ 0.5 cm.  Optical Absorption Length of gallium arsenide is ~ 0.005 cm. g. (2 pts) What is the absorption length for green light (500 nm) and infrared (900 nm) in crystalline Silicon and GaAs? Wavelength (nm) c-Si GaAs 500 10 -4 cm 10 -5 cm 900 3.3 x 10 -3 cm 0.025 cm This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

h. (2 pts) Comparing the answers for GaAs and Si from parts (f), comment on the relative solar cell thickness requirement for GaAs and Si. The absorption depth describes how deeply light penetrates into a semiconductor before being absorbed. From (f), c-Si will need thicker cell layer than GaAs solar cell. This study source was downloaded by 100000852805171 from CourseHero.com on 02-08-2024 20:03:58 GMT -06:00

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